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Quadrisecant Approximation of Hexagonal Trefoil Knot 3

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Quadrisecant Approximation of Hexagonal Trefoil Knot 3 QUADRISECANT APPROXIMATION OF HEXAGONAL TREFOIL KNOT GYO TAEK JIN AND SEOJUNG PARK Abstract. It is known that every nontrivial knot has at least two quadrise- cants. Given a knot, we mark each intersection point of each of its quadrise- cants. Replacing each subarc between two nearby marked points with a straight line segment joining them, we obtain a polygonal closed curve which we will call the quadrisecant approximation of the given knot. We show that for any hexagonal trefoil knot, there are only three quadrisecants, and the resulting quadrisecant approximation has the same knot type. 1. Preliminaries A knot is a locally flat simple closed curve in R3. Two knots said to be equivalent if there is an orientation preserving homeomorphism of R3 onto R3 carrying one to the other. The equivalence class of a knot under this equivalence relation is called its knot type. A knot is said to be nontrivial, if it does not have the knot type of a planar circle. A quadrisecant of a knot K is a straight line L such that K ∩ L has at least four components [2, 6]. Theorem 1 (Pannwitz). Every nontrivial knot has at least two quadrisecants. A polygonal knot is a knot which is the union of finitely many straight line segments. Each maximal line segment of a polygonal knot is called an edge and its end points are called vertices. Theorem 2 (Jin-Kim). The trefoil knot † can be constructed as a polygonal knot with at least six edges. arXiv:1010.2926v1 [math.GT] 14 Oct 2010 2. Quadrisecants of a Hexagonal Trefoil Knot A polygonal knot is said to be in general position if no three vertices are collinear and no four vertices are coplanar. It is clear that a quadrisecant of a polygonal knot in general position intersects the knot in finitely many points. Let K be a polygonal knot. A triangular disk ∆ determined by a pair of adjacent edges of K is said to be reducible if the ∆ intersects K only in the two edges, and irreducible otherwise. If K is a polygonal knot in general position with the least number of edges in its knot type, then every triangular disk determined by a pair of adjacent edges of K is irreducible. 2000 Mathematics Subject Classification. 57M25. Key words and phrases. knot, quadrisecant, trefoil knot, polygonal knot, quadrisecant approximation. †31 in [7] 1 2 GYOTAEKJINANDSEOJUNGPARK Let K denote a hexagonal knot with vertices v1,...,v6, and edges ei i+1 joining vi and vi+1, for i = 1,..., 6, where the subscripts are written modulo 6. For i =1,..., 6, the triangular disk with vertices at vi−1, vi, vi+1 will be denoted as ∆i. In [3], Huh and Jeon showed that in a hexagonal trefoil knot, the edges and the triangular disks intersect in a special pattern as follows: Proposition 3 (Huh-Jeon). If K is a hexagonal trefoil knot, then the triangular disks ∆1,..., ∆6 are irreducible. Furthermore, up to a cyclic relabeling of the ver- tices, the following are the only nonempty intersections among edges and triangular disks: int e23 ∩ int ∆5, int e23 ∩ int ∆6, int e45 ∩ int ∆1, int e45 ∩ int ∆2, int e61 ∩ int ∆3, int e61 ∩ int ∆4. e45 e45 e34 e34 e12 e12 e61 e61 ∆5 ∆6 e23 e23 e56 e56 Figure 1. e23 ∩ ∆5 and e23 ∩ ∆6 Theorem 4. Every hexagonal trefoil knot has exactly three quadrisecants. Proof. We first show that no three consecutive edges of K are coplanar. Suppose that three consecutive edges of K, say e12,e23 and e34, lie on a plane P . Then all vertices other than v5 and v6 lie on P . Therefore the height function of K in a normal direction to P has only one local maximum point and one local minimum point. Since such a knot is trivial, it contradicts that K is a trefoil knot. If L is a quadrisecant of K, then there are four distinct edges of K corresponding to four points of K ∩ L. If any three of these edges are consecutive along K, then they are coplanar. Therefore, by the above argument, there are only three possible sets of four edges meeting L: {e12,e23,e45,e56}, {e23,e34,e56,e61}, {e34,e45,e61,e12}. We show that there exists exactly one quadrisecant in each of the above three cases. We may assume that the vertices of K are labeled so that the edges and the triangular disks of K intersect as stated in Proposition 3. By cyclically relabeling the vertices of K, we only need to show that there exists exactly one quadrisecant meeting the edges {e12,e23,e45,e56}. Let P2 and P5 be the planes containing the triangular disks ∆2 and ∆5, respectively. We show that P2 ∩ P5 is the quadrisecant we are seeking. Notice that e45 ∩ P2 6= ∅ and e23 ∩ P5 6= ∅. Let p = e45 ∩ P2 and q = e23 ∩ P5. Then P2 ∩ P5 is the line through the two points p and q. Notice that the endpoints + − of e23 lie on the opposite sides of P5. Let P5 and P5 be the open half spaces QUADRISECANT APPROXIMATION OF HEXAGONAL TREFOIL KNOT 3 v5 v2 ☞▲ ☞▲▲ ☞ ▲ ☞ ▲ ☞ ▲ ☞ ▲ ☞ ▲ ☞ ▲ ☞ ▲ ☞ ▲ ☞ ☞ ▲ ▲▲ b ☞ q ☞ p ▲ a ☞r ☞r ▲r ▲r ☞ ☞ ▲ ▲ ☞☞ ☞ ▲ ▲ v6 ▲ v1 ☞☞ ▲ ☞ ▲ v3 v4 Figure 2. p,q,a,b ∈ P2 ∩ P5 − divided by P5 containing v2 and v3, respectively. Since v3 ∈ P5 and v4, v5 ∈ P5, − we see that ∆4 lies in P5 except along e45. Since e61 and ∆4 intersect in their − − interiors, we also know that e61 lies in P5 except at v6, hence v1 ∈ P5 . Then, we see that e12 intersects P5 in its interior. Let a = e12 ∩ P5. Then a ∈ P2 ∩ P5. + − Notice that the endpoints of e45 lie on the opposite sides of P2. Let P2 and P2 be the open half spaces divided by P2 containing v5 and v4, respectively. Similarly − as above, ∆1 lies in P2 except along e12. Since e61 and ∆3 intersect in their − − interiors, we also know that e61 lies in P2 except at v1, hence v6 ∈ P2 . Therefore we have another point b = e56 ∩ P2 in P2 ∩ P5. This completes the proof. 3. Quadrisecant Approximation Let K be a knot which has finitely many quadrisecants intersecting K in finitely many points. The intersection points cut K into finitely many subarcs. Straight- ening each subarc with its endpoints fixed, one obtains a polygonal closed curve which we will call the quadrisecant approximation of K, denoted by Kˆ . Experiments on some knots with small crossings indicate that it might be true that Kˆ is actually a knot having the knot type of K [5]. We show that the quadrise- cant approximation Kˆ of a hexagonal trefoil knot K is a trefoil knot of the same type and that the quadrisecants of Kˆ are those three of K constructed in the proof of Theorem 4. r r r r r r r r r r r r simple flipped alternating Figure 3. Three types of quadrisecants For any knot, a quadrisecant is one of the three types, simple, flipped or alternat- ing, according to the orders of the four intersection points along the quadrisecant and along the knot as indicated in Figure 3 [1]. It is easily seen from the proof of Theorem 4 and Figure 2 that the following lemma holds. 4 GYOTAEKJINANDSEOJUNGPARK Lemma 5. All quadrisecants of a hexagonal trefoil knot are alternating. Theorem 6. If K is a hexagonal trefoil knot, then its quadrisecant approximation Kˆ is also a trefoil knot that has the same knot type as K. Figure 4. Quadrisecants and quadrisecant appriximation of a hexagonal trefoil knot Proof. Let K be a hexagonal trefoil knot whose vertices and edges are labeled so that it has the intersection pattern as described in Proposition 3. Let L1,L2,L3 be the quadrisecants of K corresponding to the set of edges {e12,e23,e45,e56}, {e23,e34,e56,e61}, {e34,e45,e61,e12}, k k respectively. Let pij denote the point Lk ∩ eij and let si denote the subarc of K k which is the union of the segment of ei i+1 from vi to pi i+1 and the segment of ei−1 i k between vi and pi−1 i. We first observe the quadrisecant L1 = P2 ∩ P5 and the edges e12,e23,e45,e56. 1 1 1 1 1 k Notice that K ∩ L1 = {p12,p23,p45,p56}. We consider s2. If there is no pij ’s in the 1 ˆ 1 interior of s2, the quadrisecant approximation K has a self-intersection at p45. So, k 1 we need to show that there exists one of the pij ’s in the interior of s2. Notice that − 3 e34 lies in P5 except at v4. Thus, p34, the intersection point of L3 and e34, lies in − 3 P5 . And note that p45 lies on the plane P5. By Lemma 5, L3 is an alternating 3 3 3 3 3 quadrisecant. That is, along L3, the order of the pij ’s is p12p45p61p34. Thus, we 3 + 3 know that p12 lies in P5 . So, we know that p12 lies on the intersection of e12 and 1 the interior of s2. 1 Now, we need to show that there are no intersection points of interior of s2 and 3 k quadrisecants of K except at p12. Note that each of eij ’s has exactly two pij ’s. So, 2 it is not hard to see that p23 is the only candidate for any additional intersection 1 2 1 of s2 and quadrisecants of K.
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